Animated Solution for Physics - Kinematics: A ball released from a certain height, falls in the influence of gravity, strikes the ground and repeatedly rebounds elastically. During a time interval $t=8\text{ s}$ from the instant it is released, it covers a distance $s=20\text{ m}$. How many collisions during this time did the ball make with the ground? Acceleration of free fall is $g=10{\text{ m/s}}^2$.

Imagine you are standing on a balcony, holding a perfectly elastic ball. You let it go. It falls, strikes the ground, and bounces back up to the exact same height from which you dropped it. This mesmerizing, repetitive motion is a classic setup in kinematics, but when you are given only the total time and the total distance covered, the problem suddenly transforms into a beautiful mathematical puzzle.

I know this setup might look a bit intimidating at first glance because the ball is constantly changing direction. But let's take a breath. The secret to cracking this problem lies in breaking the complex motion into simple, identical building blocks.

The "One-Way Trip"

Instead of looking at the entire chaotic bouncing sequence, let's focus on just one segment of the motion. We will call this a "one-way trip". A one-way trip is either the ball falling from its maximum height $h$ to the ground, or the ball rising from the ground back to its maximum height $h$.

Let the time taken for one such trip be $t_0$. Using the second equation of motion for a body starting from rest (at the top of its trajectory), we can write the distance of one trip as:

Every single trip, whether up or down, covers this exact distance $h$ in this exact time $t_0$.

Now, let's zoom out. Suppose the ball completes $k$ such one-way trips in the total given time $t$. If $k$ is a perfect integer, then the total distance $s$ is simply the number of trips multiplied by the distance of one trip:

Similarly, the total time $t$ is the number of trips multiplied by the time of one trip:

Here is where the magic happens. Even if $k$ is not a perfect integer (meaning the ball is caught mid-air at the end of the 8 seconds), this relationship serves as an incredibly accurate approximation. Why? Because the average speed of the ball over any one-way trip is constant!

Let's substitute our expressions for $h$ and $t_0$ into the total distance equation. We know that $t_0=\frac{t}{k}$. Plugging this into our equation for $h$:

Simplifying this expression, we get:

Rearranging to solve for $k$, the total number of one-way trips, we arrive at our master equation:

Let's plug in the numbers given in the problem: $t=8\text{ s}$, $s=20\text{ m}$, and $g=10{\text{ m/s}}^2$.

Wow! The math worked out perfectly. The ball makes exactly 16 one-way trips.

Trips vs. Collisions

We aren't done yet. The question asks for the number of collisions, not trips. We need to map our trips to collisions.

Let's trace the ball's journey: - Trip 1: Falling down $\to$ 1st Collision - Trip 2: Bouncing up - Trip 3: Falling down $\to$ 2nd Collision - Trip 4: Bouncing up - Trip 5: Falling down $\to$ 3rd Collision

Do you see the pattern? The collisions happen at the end of the 1st, 3rd, 5th, 7th... trips. In mathematical terms, the $n$-th collision occurs exactly at the end of the $(2n-1)$-th trip.

Since our ball completes 16 trips, we can set up a simple inequality to find the maximum number of collisions $n$:

Since a physical collision is a discrete event, the number of collisions must be an integer. Therefore, we take the greatest integer less than or equal to 8.5.

The Intuitive Approach

If you love physics intuition, here is another way to think about it. The average speed of the ball over the entire 8 seconds is $v_{avg}=\frac{s}{t}=\frac{20}{8}=2.5\text{ m/s}$.

Because the motion is perfectly repetitive, the average speed of a single one-way trip must also be $2.5\text{ m/s}$. The average speed of a falling object is $\frac{h}{t_0}=\frac{h}{\sqrt{2h/g}}=\sqrt{\frac{gh}{2}}$.

Equating the two: $\sqrt{\frac{10h}{2}}=2.5$, which gives $h=1.25\text{ m}$.

The time for one trip is $t_0=\sqrt{\frac{2(1.25)}{10}}=0.5\text{ s}$. In 8 seconds, there are exactly $\frac{8}{0.5}=16$ trips. Half of these trips end in a collision, giving us exactly 8 collisions. Physics is beautiful when it all connects!